My question is as in the title. That is, consider solving the equation $x^2-6=0$ in $\mathbb{R}$ and in the 5-adic field $\mathbb{Q}_5$ respectively. We obtain one $\sqrt{6}\in\mathbb{R}$ and one $\sqrt{6}\in\mathbb{Q}_5$. Could you tell me whether the real $\sqrt{6}$ and the 5-adic number $\sqrt{6}$ are equal to each other? Thanks very much!

$\begingroup$What do you mean by "equal" in this case?$\endgroup$
– 5xumJul 27 '15 at 14:14

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$\begingroup$To meaningfully compare them you would need a universe $\Omega$ and injective mappings $\Bbb{R}\to\Omega$ and $\Bbb{Q}_5\to\Omega$. It is known that $\Bbb{C}$ contains a copy of $\Bbb{Q}_5$, but nobody (AFAICT) can describe that injective mapping. Mind you, with $\Omega=\Bbb{C}$ the only possible ambiguity is with the sign. But, to reiterate, without that injective mapping from the $5$-adics any comparison is meaningless.$\endgroup$
– Jyrki LahtonenJul 27 '15 at 14:17

$\begingroup$@5xum by "equal" I mean: if the real $\sqrt{6}$ can be expressed as a formal series $a_{-N}p^{-N}+\ldots+a_0+a_1 p+a_2 p^2+\ldots$, then it is the same as that of the 5-adic number $\sqrt{6}$, just as every rational number $\dfrac{p}{q}$.$\endgroup$
– C.C.Jul 27 '15 at 14:40

$\begingroup$@JyrkiLahtonen Thank you very much! What do you mean by $\mathbb{C}$ contains a copy of $\mathbb{Q}_5$? Here I avoid the ambiguity caused by the sign, i.e. I only consider the positive root $\sqrt{6}$ in $\mathbb{R}$ and the root 5-adically closer to 4 in $\mathbb{5}$. Maybe my question could be modified as "can the quadratic field $\mathbb{Q}(\sqrt{6})$ be embedded into $\mathbb{Q}_5$"?$\endgroup$
– C.C.Jul 27 '15 at 14:52

$\begingroup$@5xum Thank you very much! But I cannot understand "Because in the reals, if such an infinite series would only converge if it is finite". Could you please explain it a bit more?$\endgroup$
– C.C.Jul 27 '15 at 14:59

1 Answer
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The question of whether two "numbers" are "equal" is a somewhat subtle one.

For example, lets work with a simpler number, namely "2".

Certainly $2\in\mathbb{Z}$, but also $2\in\mathbb{Q},2\in\mathbb{R}$. Of course, they're all called the same name, and they satisfy some of the same properties: For example, in all three situations, $2 = 1+1$, and indeed one would often say that all three 2's are "the same".

However, in $\mathbb{Z}$, there is no solution to the equation $2x = 1$, whereas there is in $\mathbb{Q}$ and $\mathbb{R}$, namely, "1/2".

Similarly, in $\mathbb{Q}$, 2 does not have a square root, whereas it does in $\mathbb{R}$.

Does this mean that $2\in\mathbb{Z},2\in\mathbb{Q},2\in\mathbb{R}$ are all different?

As another example, in $\mathbb{Z}/4\mathbb{Z}$, it's common to label the elements $\{0,1,2,3\}$, then $2\in\mathbb{Z}/4\mathbb{Z}$ satisfies the equation "2+2+2 = 2", which of course is not satisfied in $\mathbb{Z},\mathbb{Q}$, or $\mathbb{R}$.

At this point, one is led to consider - "What do we mean by 2?"

After all, the symbol "2" is simply an arabic numeral. By itself it has no meaning, until we associate to the symbol an idea. In this case, often the idea associated to "2" is the sum $1+1$, where 1 is the multiplicative identity in your ring. By the definition of a ring (with unity), such a multiplicative identity exists, and we can always add two elements of a ring to get an element of the ring, and thus $1+1$ exists as an element in the ring, which we call "2".

POINT: Symbols like "$2$", or "$\sqrt{2}$" are just symbols - they are labels that point to ideas we have in our mind. It doesn't really make sense to compare labels. For example, we could have also chosen to use "2" to refer to the number $1+1+1\in\mathbb{Z}$. Would that mean that suddenly $1+1 = 1+1+1\in\mathbb{Z}$? Of course not.

Thus, instead of comparing labels, we should instead be comparing "ideas". Why is it reasonable to say that $2\in\mathbb{Z}$ is equal to $2\in\mathbb{Q}$? On the other hand, $2\in\mathbb{Z}$ is not "equal" to $2\in\mathbb{Z}/4\mathbb{Z}$ - if they were equal, then why is $2+2+2=2$ only valid in $\mathbb{Z}/4\mathbb{Z}$, but not $\mathbb{Z}$?

The answer of course, is that in the first case there is an injective ring homomorphism $\mathbb{Z}\hookrightarrow\mathbb{Q}$, thus allowing us to think of $\mathbb{Z}$ as a subring of $\mathbb{Q}$, without losing any information, whereas of course there is no such injection between $\mathbb{Z}/4\mathbb{Z}$ and $\mathbb{Z}$.

Thus, when we ask: Is $2\in\mathbb{Z}$ equal to $2\in\mathbb{Q}$, what we really mean is: Is the image of $2\in\mathbb{Z}$ under the injection $\mathbb{Z}\hookrightarrow\mathbb{Q}$ the same as $2\in\mathbb{Q}$?

POINT: You can only ask if two objects $x,y$ are equal, if $x,y$ belong to the same set. If $x,y$ do not belong to the same set, then you must find some way to think about them as both lying in the same set. For example, let's consider the rings $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}[x]/(x^2-2)$. Is $\sqrt{2}$ in $\mathbb{Q}(\sqrt{2})$ equal to $x\in\mathbb{Q}[x]/(x^2-2)$? At first, the two objects clearly have different "labels" ("$\sqrt{2}$" vs "$x$"). But on the other hand, we can view $\mathbb{Q}(\sqrt{2})$ as lying in $\mathbb{R}$, and similarly we can inject $\mathbb{Q}[x]/(x^2-2)$ into $\mathbb{R}$ by sending $x\mapsto \sqrt{2}$, at which point we see that $\sqrt{2}\in\mathbb{Q}(\sqrt{2})$ is "equal" to $x\in\mathbb{Q}[x]/(x^2-2)$ relative to these injections. Note that we could have just as easily chosen the injection $\mathbb{Q}[x]/(x^2-2)\hookrightarrow\mathbb{R}$ sending $x\mapsto -\sqrt{2}$, at which point instead we would get that $x\in\mathbb{Q}[x]/(x^2-2)$ is equal to $-\sqrt{2}$

Finally, coming to your original question: To be able to compare $\sqrt{6}\in\mathbb{R}$ and $\sqrt{6}\in\mathbb{Q}_5$, one must first find a "larger" field $K$ and injections $\mathbb{R}\hookrightarrow K$ and $\mathbb{Q}_5\hookrightarrow K$.

It's known that there exist such injections if $K = \mathbb{C}$, but it's impossible to "describe" the injections fully in a finite number of symbols. Though, even here, what do you mean by $\sqrt{6}\in\mathbb{Q}_5$? Note that $\mathbb{Q}_5$ is not an ordered field (mainly because it contains a square root of $-1$), and thus you cannot simply say that $\sqrt{6}$ is the "positive" solution to $X^2-6$. However, it is safe to say that under the injections $\mathbb{R}\hookrightarrow \mathbb{C}$ and $\mathbb{Q}_5\hookrightarrow\mathbb{C}$, the image of $\sqrt{6}\in\mathbb{R}$ in $\mathbb{C}$ is equal to the image of ONE of the (two) solutions of $X^2-6$ over $\mathbb{Q}_5$ in $\mathbb{C}$.

EDIT: To address your question in the comments, note that there is a difference between "formal laurent series" (laurent = power series where you allow finitely many negative power terms) and "laurent series". The former refers to an element of a ring of the form $R((X))$, whereas the latter refers to the limit of the partial sums $\lim_{n\rightarrow\infty}\sum_{i=k}^n a_ip^i$, $k\in\mathbb{Z}$.

Note that $\mathbb{Q}_5$ is NOT a formal laurent series ring. However, it is true that every 5-adic number is a limit of partial sums of the form $\sum_{i=k}^n a_ip^i$, but to even say this, you need to specify a topology/metric, which $p$-adic numbers are naturally equipped with. However, this topology is wildly different from the topology on $\mathbb{R}$ or $\mathbb{C}$. As a result, there is no continuous injection $\mathbb{Q}_5\hookrightarrow\mathbb{C}$, and hence no way to express $\sqrt{6}$ as a laurent series in powers of 5 in $\mathbb{C}$.

$\begingroup$Wow. You wrote a book! At least it looks that way on smart phone. Nice story. Vote up.$\endgroup$
– johannesvalksJul 27 '15 at 18:12

$\begingroup$@oxeimon Thanks very much for your very nice explanations! But I am a little confused about "As a result, $\ldots$, and hence no way to express $\sqrt{6}$ as a laurent series in powers of 5 in $\mathbb{C}$", since according to your description, $\sqrt{6}$ is a laurent series rather than a formal laurent series? Also, could you please give an explicit embedding as stated in EDIT 2? Moreover, can the same thing be said to a number field $\mathcal{k}$, say $\mathbb{Q}(\sqrt{5})$?$\endgroup$
– C.C.Jul 28 '15 at 6:53

$\begingroup$@oxeimon That is, if $\mathcal{v}$ is a finite place of $\mathcal{k}$, does there exist a embedding $\mathcal{k}(\sqrt{2})\hookrightarrow \mathcal{k}_{\mathcal{v}}$?$\endgroup$
– C.C.Jul 28 '15 at 6:53

$\begingroup$@C.Christopher Well just try to write $\sqrt{6}$ as something of the form $\sum_{i=k}^\infty a_i 5^i$, where $k\in\mathbb{Z}$. Note that this will only converge in the usual topology on $\mathbb{C}$ if there are only finitely many nonzero $a_i$'s. What do you mean by explicit? You just send $\sqrt{6}\in\mathbb{Q}(\sqrt{6})$ to the laurent series in $\mathbb{Q}_5$, which you can calculate by using Hensel's lemma.$\endgroup$
– oxeimonJul 28 '15 at 8:06

$\begingroup$There is an embedding $k(\sqrt{2})\hookrightarrow k_v$ if and only if $k_v$ has a square root of 2.... The existence of such a square root is described by Hensel's lemma.$\endgroup$
– oxeimonJul 28 '15 at 8:08